Answer :

Let p(x) = x^{3} – 3√5x^{2} – 5x + 15√5

As p(x) is cubic polynomial it will have 3 zeroes

As (x – √5) is a factor of p(x), p(√5) will be 0 by remainder theorem hence one zero of p(x) is √5

Divide p(x) by (x – √5)

Using dividend = divisor × quotient + remainder

⇒ p(x) = (x – √5)(x^{2} – 2√5x – 15) + 0

Remaining two zeroes of p(x) we can get by equating x^{2} – 2√5x – 15 = 0

⇒ x^{2} – 2√5x – 15 = 0

⇒ x^{2} – 3√5x + √5x – 15 = 0

⇒ x(x – 3√5) + √5(x – 3√5) = 0

⇒ (x + √5)(x – 3√5) = 0

⇒ x + √5 = 0 and x – 3√5 = 0

⇒ x = –√5 and x = 3√5

Hence all zeroes of x^{3} – 3√5x^{2} – 5x + 15√5 are x = √5, x = –√5 and x = 3√5

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